The Stacks project

Proof. This follows formally from the fact that $f^{-1}$ is an exact left adjoint of $f_*$, see Homology, Lemma 12.29.1. $\square$

Proof. Part (1) follows Categories, Lemma 4.35.15 as $(\mathit{Sch}/S)_{fppf}$ has fibre products.

Let $a, b : x \to y$ be morphisms of $\mathcal{X}$. Set $U = p(x)$ and $V = p(y)$. The category of schemes has equalizers hence we can let $W \to U$ be the equalizer of $p(a)$ and $p(b)$. Denote $c : z \to x$ a morphism of $\mathcal{X}$ lying over $W \to U$. The equalizer of $a$ and $b$, if it exists, is the equalizer of $a \circ c$ and $b \circ c$. Thus we may assume that $p(a) = p(b) = f : U \to V$. As $\mathcal{X}$ is fibred in groupoids, there exists a unique automorphism $i : x \to x$ in the fibre category of $\mathcal{X}$ over $U$ such that $a \circ i = b$. Again the equalizer of $a$ and $b$ is the equalizer of $\text{id}_ x$ and $i$. Recall that the $\mathit{Isom}_\mathcal {X}(x)$ is the presheaf on $(\mathit{Sch}/U)_{fppf}$ which to $T/U$ associates the set of automorphisms of $x|_ T$ in the fibre category of $\mathcal{X}$ over $T$, see Stacks, Definition 8.2.2. If $\mathit{Isom}_\mathcal {X}(x)$ is representable by an algebraic space $G \to U$, then we see that $\text{id}_ x$ and $i$ define morphisms $e, i : U \to G$ over $U$. Set $M = U \times _{e, G, i} U$, which by Morphisms of Spaces, Lemma 67.4.7 is a scheme. Then it is clear that $x|_ M \to x$ is the equalizer of the maps $\text{id}_ x$ and $i$ in $\mathcal{X}$. This proves (2).

If $\mathcal{X} = [U/R]$ for some groupoid in algebraic spaces $(U, R, s, t, c)$ over $S$, then the hypothesis of (2) holds by Bootstrap, Lemma 80.11.5. If $\mathcal{X}$ is an algebraic stack, then we can choose a presentation $[U/R] \cong \mathcal{X}$ by Algebraic Stacks, Lemma 94.16.2. $\square$

Proof. By Categories, Lemma 4.35.15 we see that a fibre product in $\mathcal{X}$ is any commutative square lying over a fibre product diagram in $(\mathit{Sch}/S)_{fppf}$. Similarly for $\mathcal{Y}$. Hence (1) is clear.

Let $x \to x'$ be the equalizer of two morphisms $a, b : x' \to x''$ in $\mathcal{X}$. We will show that $f(x) \to f(x')$ is the equalizer of $f(a)$ and $f(b)$. Let $y \to f(x)$ be a morphism of $\mathcal{Y}$ equalizing $f(a)$ and $f(b)$. Say $x, x', x''$ lie over the schemes $U, U', U''$ and $y$ lies over $V$. Denote $h : V \to U'$ the image of $y \to f(x)$ in the category of schemes. The morphism $y \to f(x)$ is isomorphic to $f(h^*x') \to f(x')$ by the axioms of fibred categories. Hence, as $f$ is faithful, we see that $h^*x' \to x'$ equalizes $a$ and $b$. Thus we obtain a unique morphism $h^*x' \to x$ whose image $y = f(h^*x') \to f(x)$ is the desired morphism in $\mathcal{Y}$. $\square$

Proof. We think of objects in $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$ as quadruples $(U, x, z, \alpha )$ where $\alpha : f(x) \to g(z)$ is an isomorphism over $U$, see Categories, Lemma 4.32.3. A morphism $(U, x, z, \alpha ) \to (U', x', z', \alpha ')$ is a pair of morphisms $a : x \to x'$ and $b : z \to z'$ compatible with $\alpha $ and $\alpha '$. Thus it is clear that if $f$ and $g$ are faithful, so is the functor $\mathcal{X} \times _\mathcal {Y} \mathcal{Z} \to \mathcal{Y}$. Now, suppose that $(a, b), (a', b') : (U, x, z, \alpha ) \to (U', x', z', \alpha ')$ are two morphisms of the $2$-fibre product. Then consider the equalizer $x'' \to x$ of $a$ and $a'$ and the equalizer $z'' \to z$ of $b$ and $b'$. Since $f$ commutes with equalizers (by Lemma 96.17.3) we see that $f(x'') \to f(x)$ is the equalizer of $f(a)$ and $f(a')$. Similarly, $g(z'') \to g(z)$ is the equalizer of $g(b)$ and $g(b')$. Picture

It is clear that the dotted arrow exists and is an isomorphism. However, it is not a priori the case that the image of $\alpha ''$ in the category of schemes is the identity of its source. On the other hand, the existence of $\alpha ''$ means that we can assume that $x''$ and $z''$ are defined over the same scheme and that the morphisms $x'' \to x$ and $z'' \to z$ have the same image in the category of schemes. Redoing the diagram above we see that the dotted arrow now does project to an identity morphism and we win. Some details omitted. $\square$

Proof. By Stacks, Lemma 8.10.3 the functor $f$ is continuous and cocontinuous. Hence by Modules on Sites, Lemma 18.16.2 the functor $f^{-1} : \textit{Ab}(\mathcal{Y}_\tau ) \to \textit{Ab}(\mathcal{X}_\tau )$ has a left adjoint $f_! : \textit{Ab}(\mathcal{X}_\tau ) \to \textit{Ab}(\mathcal{Y}_\tau )$. To see (1) we apply Modules on Sites, Lemma 18.16.3 and to see that the hypotheses of that lemma are satisfied use Lemmas 96.17.2 and 96.17.3 above. Part (2) follows from this formally, see Homology, Lemma 12.29.1. $\square$

Proof. Recall that $f$ is a continuous and cocontinuous functor of sites and that $f^{-1}\mathcal{O}_\mathcal {Y} = \mathcal{O}_\mathcal {X}$. Hence Modules on Sites, Lemma 18.41.1 implies $f^*$ has a left adjoint $f_!^{Mod}$. Let $x$ be an object of $\mathcal{X}$ lying over the scheme $U$. Then $f$ induces an equivalence of ringed sites

as both sides are equivalent to $(\mathit{Sch}/U)_\tau $, see Lemma 96.9.4. Modules on Sites, Remark 18.41.2 shows that $f_!$ agrees with the functor on abelian sheaves.

Assume now that $\mathcal{X}$ has equalizers and that $f$ is faithful. Lemma 96.17.5 tells us that $f_!$ is exact. Finally, Homology, Lemma 12.29.1 implies the statement on pullbacks of injective modules. $\square$